「THUPC 2017」小 L 的计算题 / Sum [生成函数 + 多项式Ln]

最新推荐文章于 2021-07-23 14:06:04 发布

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最新推荐文章于 2021-07-23 14:06:04 发布

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分类专栏：多项式

本文链接：https://blog.youkuaiyun.com/sslz_fsy/article/details/98896634

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本文深入探讨了多项式运算的高效实现，包括分治FFT算法及其在多项式求逆和对数运算中的应用。通过具体代码示例，详细解析了如何使用FFT进行多项式的快速乘法和求逆，以及如何利用多项式对数解决复杂问题。

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传送门

直接构造答案的生成函数

$F(x)=\sum_i f_i * x ^i=\sum_{i}\sum_{j}a_j^i*x ^i=\sum_j\sum_i(a_j*x)^i= \sum_j \frac{1} {1-a_j* x}=n-\sum_j\frac{-a_j*x}{1-a_j*x}=n-x\sum_j \frac{-a_j}{1-a_j*x}=n-x\sum_j (ln(1-a_j*x))'= n- x (ln\prod_j (1-a_j*x))'$

然后可以分治 FFT + 多项式 Ln

#include<bits/stdc++.h>
#define N 1600050
using namespace std;
typedef long long ll;
int read(){
	int cnt = 0, f = 1; char ch = 0;
	while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1;}
	while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
	return cnt * f;
}
const int Mod = 998244353, G = 3;
int T, n; ll a[N], A[N], B[N], C[N], D[N], F[N], f[N], g[N];
int up, bit, rev[N];
ll add(ll a, ll b){ return (a + b) % Mod;}
ll mul(ll a, ll b){ return a * b % Mod;}
ll power(ll a, ll b){ ll ans = 1;
	for(;b;b>>=1){ if(b&1) ans = mul(ans, a); a = mul(a, a);}
	return ans;
}
void Init(int len){ len <<= 1; up = 1; bit = 0;
	while(up < len) up <<= 1, bit++; 
	for(int i = 0; i < up; i++) rev[i] = (rev[i>>1] >> 1) | ((i&1) << (bit-1));
}
void NTT(ll *a, int flag){
	for(int i = 0; i < up; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(int i = 1; i < up; i <<= 1){
		ll wn = power(G, (Mod - 1) / (i << 1));
		if(flag == -1) wn = power(wn, Mod - 2);
		for(int j = 0; j < up; j += (i<<1)){
			ll w = 1;
			for(int k = 0; k < i; k ++, w = mul(w, wn)){
				ll x = a[k + j], y = mul(w, a[k + j + i]);
				a[k + j] = add(x, y); a[k + j + i] = add(x, Mod - y);
			}
		}
	}
	if(flag == -1){
		ll inv = power(up, Mod - 2);
		for(int i = 0; i < up; i++) a[i] = mul(a[i], inv);
	}
} 
void Solve(int l, int r, int len){
	if(l == r){ a[l] = Mod-a[l]; return;}
	int mid = (l+r) >> 1;
	Solve(l, mid, len >> 1);
	Solve(mid+1, r, len >> 1);
	Init(len >> 1);
	for(int i = 0; i < up; i++) A[i] = B[i] = 0;
	A[0] = B[0] = 1;
	for(int i = l; i <= mid; i++) A[i - l + 1] = a[i];
	for(int i = mid+1; i <= r; i++) B[i - mid] = a[i];
	NTT(A, 1); NTT(B, 1);
	for(int i = 0; i < up; i++) A[i] = mul(A[i], B[i]);
	NTT(A, -1);
	for(int i = l; i <= r; i++) a[i] = A[i - l + 1];
}
void Inv(ll *a, ll *b, int len){
	if(len == 1){ b[0] = power(a[0], Mod - 2); return;}
	Inv(a, b, (len + 1) >> 1);
	Init(len);
	for(int i = 0; i < len; i++) F[i] = a[i];
	for(int i = len; i < up; i++) F[i] = 0;
	NTT(F, 1); NTT(b, 1);
	for(int i = 0; i < up; i++) b[i] = mul(b[i], Mod + 2 - mul(b[i], F[i]));
	NTT(b, -1);
	for(int i = len; i < up; i++) b[i] = 0;
}
void Ln(ll *a, ll *b, int len){
	for(int i = 1; i < len; i++) C[i-1] = mul(a[i], i); C[len - 1] = 0;
	Inv(a, D, len);
	Init(len);
	NTT(C, 1); NTT(D, 1);
	for(int i = 0; i < up; i++) C[i] = mul(C[i], D[i]);
	NTT(C, -1);
	for(int i = 1; i < up; i++) b[i] = mul(C[i-1], power(i, Mod-2));
	b[0] = 0;
	for(int i = 0; i < up; i++) C[i] = D[i] = 0;
}
int main(){
	T = read();
	while(T--){
		memset(a, 0, sizeof(a));
		n = read(); a[0] = 1;
		for(int i = 1; i <= n; i++) a[i] = read();
		int len = 1; while(len < (n + 1) << 1) len <<= 1;
		Solve(1, n, len); Ln(a, f, n + 1);
		for(int i = 1; i <= n + 1; i++) g[i-1] = mul(f[i], i);
		for(int i = n; i >= 1; i--) g[i] = Mod - g[i-1]; 
		ll ans = 0;
		for(int i = 1; i <= n; i++) ans ^= g[i];
		cout << ans << "\n";
	} return 0;
}